Normal subgroup and Lie algebra I have an exercise of Lie group as follows: "Let $G,H$ be closed connected subgroup of $GL_n(\mathbb{R})$, and $H$ be subgoup of $G$. Suppose that $\operatorname{Lie}(H)$ is an ideal of $\operatorname{Lie}(G)$. Prove that $H$ is a normal subgroup of $G$."
I get stuck to solve this problem. Also I have no idea to use the connectedness of $G$ and $H$. Some one can help me? Thanks a lot!
 A: This is essentially an application of the Lie subalgebra-subgroup correspondence:

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Suppose that $\mathfrak{h}$ is a Lie subalgebra of $\mathfrak{g}$. Then there is a unique connected immersed Lie subgroup $H\subseteq G$ whose Lie algebra corresponds to $\mathfrak{h}$. 

You are given $H$ a closed connected Lie subgroup of the Lie group $G$. Choose $g\in G$, and let $H' = gHg^{-1}$. Then $H'$ is a Lie group with corresponding Lie algebra $Lie(H')\subseteq Lie(G)$. However, the assumption that $Lie(H)$ is an ideal of $Lie(G)$ says exactly that $Lie(H') = Lie(H)$. You then have that $H$ and $H'$ are two connected Lie subgroups of $G$ with the same Lie algebra. By the uniqueness in the above theorem, it follows that $H = H'$, and hence that $H$ is normal. 
A: There seems to be a full proof contained in lemma 0.1 here http://math.berkeley.edu/~ianagol/261A.F09/Simplegroups.pdf
Here is the argument:
Claim: For $X \in \frak{g}$, $Y \in \frak{h}$, we have $e^Xe^Ye^{-X}\in H$.
Proof of claim:  Denote by $\operatorname{exp}$ the exponential map $\operatorname{End}(\frak{g}) \rightarrow $ $\operatorname{Aut}(\mathfrak{g}) $ (where Aut and End are the vector space automorphisms and endomorphisms) .   Since $\mathfrak{h}$ is an ideal, $\operatorname{ad}_X^n$ preserves $\mathfrak{h}$ for all $n \in \mathbb{N}$, and therefore so does $\operatorname{exp}(\operatorname{ad}_X)$.  We therefore have $$e^Xe^Ye^{-X}=e^{\operatorname{Ad}_{e^X}(Y)}=e^{\operatorname{exp}(\operatorname{ad}_X)Y} \in e^{\mathfrak{h}} $$
This establishes the claim.
Now the fact that $G$ normalizes $H$ follows from the fact that $$H= \bigcup_{n\in \mathbb{N}} (e^\mathfrak{h})^n$$ $$G= \bigcup_{n \in \mathbb{N}}(e^\mathfrak{g})^n$$ Since $G$ and $H$ are connected.
In particular, this doesn't seem to use the fact that $G$ is a closed subgroup of $GL(n, \mathbb{R})$, although this is a hypothesis of lemma 0.1. Nor does it require even that $H$ be closed in $G$.
A: I will two theorems that can be found in Lie groups and Algebraic group book by Vinberg. 
$Theorem:1$ A a homomorphism from a connected lie group H to a lie group G is 
uniquely determined by tangent algebra homomorphism. 
$Theorem:2$
Let $f:H \rightarrow G$ where is H is connected.If $G_1\subset G$ such that 
$df(Lie(H))\subset Lie(G_1)$. Then $f(H)\subset G_1$. 
Let take arbitary $g\in G$ then $a(g):x\rightarrow gxg^{-1}$. The differential of this map say Ad(g) coincide with the adjoint representation of lie algebra hence this map Lie(H) to Lie(H). So $gHg^{-1}\subset H$ for all g hence H is normal.//
You can find all the proofs in the book I mentioned in chapter 1 section 2 I hope. 
